User kristal cantwell - MathOverflow

Answer by Kristal Cantwell for Is rigour just a ritual that most mathematicians wish to get rid of if they could?

2013-04-26T02:59:24Z

There is the case of Hilbert's 16th problem in which errors were found in some proofs. See "Centennial History of Hilbert's 16th Problem", Yu. Ilyashenko, Bull. Amer. Math. Soc. 39 (2002), 301-354. One result which was published in 1923 was found to be faulty in 1981 over 50 years later.

There is another thread about rigor in mathematics which has some examples: http://mathoverflow.net/questions/37610/demonstrating-that-rigour-is-important

Answer by Kristal Cantwell for Have we ever lost any mathematics?

2012-05-10T04:19:48Z

There are "Lectures on Lost Mathematics" by B. Grünbaum. They were given at the University of Washington in 1975. The notes are available here

Answer by Kristal Cantwell for Great mathematical figures and/or diagrams?

2012-12-14T20:35:11Z

H.S.M. Coxeter's books include Regular Polytopes. This book deals with the classification of regular polytopes. In this book there are Coxeter diagrams which are closely related to Dynkin diagrams. In his works are many diagrams,figures and illustrations. They influenced M.C Escher. Many of Escher's works reflect his ideas. In 1996 Coxeter published a paper on one of these "Circle Limit III." For more information see here and here.

Answer by Kristal Cantwell for Real-world applications of mathematics, by arxiv subject area?

2012-01-18T21:02:55Z

math.GT Geometric Topology: used for topological quantum field theory (TQFT). There is information on applications of this subject area: here.

Answer by Kristal Cantwell for What makes four dimensions special?

2012-05-11T16:29:19Z

Here is more about regular polytopes and four dimensions. For regular convex polytopes there are six regular polytopes and then for all dimensions higher than 4 there are three. For nonconvex regular polytopes there are 10 in four dimensions and zero in all higher dimensions. For convex euclidean tessellations there are 3 in four dimensions zero in all higher dimensions.

However there are some cases in which the 5th and 6th dimension have different values: For convex hyperbolic tessellations 4 in the fourth dimension, 5 in the fifth and zero in all higher dimensions. For nonconvex hyperbolic tessellations there are zero in four dimensions four in the fifth dimension and zero in all higher dimensions.

I used the wikipedia article "List of regular polytopes" which is available here as a source for the above information.

Answer by Kristal Cantwell for Interesting conjectures "discovered" by computers and proved by humans?

2012-03-25T18:37:07Z

There is a related question on experimental mathematics here.

Answer by Kristal Cantwell for Action of Isometries on a Line in the Plane

2012-02-22T00:24:50Z

You seem to be missing glide reflections. For any translation along a line you can reflect across the line getting a glide reflection. They are in the wikipedia article you sited.

Answer by Kristal Cantwell for Vertex-transitive polytopes in any dimension with any number of vertices?

2012-02-21T21:33:03Z

There is no three dimensional vertex transitive polyhedron with 7 vertices. The rotation groups of three dimensional polyhedra are finite quaternion groups. The only finite quaterionion groups whose order is divisible by 7 are associated with polygons with 7 or a multiple of 7 sides. So there have to be another element in the group to move points out of the plane and this result in more than 7 points. There is a complete classification of finite isometric groups that leave one point fixed in dimensions 3 and 4.

There are also no three dimensional vertex transitive polyhedra with 5 vertices but in that case there are more finite quaternion groups whose order is divisible by 5. When these are examined there are no three dimensional vertex transitive polyhedra but there is one in four dimensions namely the four dimensional simplex.

Answer by Kristal Cantwell for A "couple" of questions on Gauss's mathematical diary

2012-02-21T01:42:14Z

I believe I have a reference for Schuhmann:

Schuhmann E. 1976 Vicimus GEGAN, Interpretationsvarianten zu einer Tagebuchnotiz von C.F. Gauss, Naturwiss. Tech. Medezin. 13.2, 17-20.

Answer by Kristal Cantwell for The missing Euler Idoneal numbers

2010-04-11T03:13:25Z

The paper you have quoted says that if the generalized Riemann hypothesis holds then there are only 65 idoneal numbers(see corollary 23). This agrees with the first comment to your answer. According to the paper the error made that allowed only one more idoneal number is the assumption that it has been proved that there are no idoneal even numbers greater than 1848(see remark 24). So if the generalized Riemann hypothesis does not hold that does not imply that there are additional idoneal numbers. Even if there were more there still could be one more odd idoneal number. It is only if the idoneal number after 1848 is even that 4 times that number will also be idoneal.

Answer by Kristal Cantwell for Counterexamples in PDE

2011-06-26T05:27:14Z

In the paper "One cannot hear the shape of a drum", Carolyn Gordon, David L. Webb and Scott Wolpert give an example of two simply connected regions which are isospectral but not isometric. The article is available here. It was in answer to the article "Can you hear the shape of a drum by Mark Kac. Also see this article for more information about this problem.

Answer by Kristal Cantwell for Which Fibonacci numbers are the sum of two squares?

2011-06-23T20:44:01Z

If we assume the conjecture that 6 is the only even number such that F_2n is the sum of two squares then 2 cannot divide n. Then we also have F_2n=F_n*L_n and so if F_2n is the sum of two squares F_n is the sum of two squares since n is odd so F_2n is the sum of two squares if L_n is the sum of two squares. Also if N is odd and F_2N is the sum of two squares then L_n is the sum of two squares. So if we assume the question is equivalent to the following are there an infinite number of odd n such that L_n is the sum of two squares.

Answer by Kristal Cantwell for Rolling-ball game

2011-05-13T05:39:18Z

I can show the game is finite which answers one of the questions in the comments and shows the answer to question two is not infinity. The arcs in on the sphere from the rolling are sections of circles of unit radius the angles between the arcs are either 90 or zero. Furthermore there is finite limit to the number of rotations without switching the direction of rotation. So for a large number of rotations there will be a set of arcs with a large number of 90 degree angles. But this will mean a large angle deficiency and an arbitrarily large area but there is a finite limit to the area of subset of the surface of the sphere so there is a contradiction.

Answer by Kristal Cantwell for Most striking applications of category theory?

2010-03-25T17:43:09Z

The finite vector space analog to Ramsey's theorem was proved using categories the paper is available here

Answer by Kristal Cantwell for What is the "Physically Consistent" proper subset of arithmetic?

2011-03-19T00:08:20Z

Presburger arithmetic which is the first order theory of natural numbers with addition has been proven to be consistent by Mojżesz Presburger. My reference for this is the wikipedia article on Presburger arithmetic.

Answer by Kristal Cantwell for ask for the time complexity of a convex quadratically constrained quadratic program (QCQP) problem

2011-03-14T18:30:45Z

According to the Wikipedia article at http://en.wikipedia.org/wiki/NP-hard it is NP-hard. The Wikipedia article gives as a reference a book which is available at http://www.stanford.edu/~boyd./cvxbook/

Answer by Kristal Cantwell for Cops and drunken robbers

2011-03-13T19:51:10Z

I think that I have an example where the optimal strategy for a random robber is different from the normal winning strategy.

Let me specify the graph first. We have 5 points A,B,C,D and E forming a cycle. So an edge connects A to B, and an edge connects B to C etc. points A and C are also connected by an edge. We also add a billion points connected to only D and a billion points connected to only E.

Now let me specify the position. The cop is on vertex A and the robber is on vertex D.

Now the winning strategy is for the cop to move to vertex B then If the robber moves to E then The cop moves to C and then if the robber moves back to D The cop moves to D and we are done. If then instead of moving to D the robber moves to one of the many points connected only to E the cop moves to A and now the robber must move to E and then the cop captures the robber there.

Now if the robber's moves are random The optimal strategy is for the cop to move to vertex C because the robber will with odds a billion to one move that the robber will move to one of the points adjacent only to D. Then the cop will move to D and in the next move the robber must move into D and and the cop will catch the robber after only two moves. Then if the robber moves to E the cop moves to A and again the odds are a billion to one that the robber will move to a point adjacent only to E and the cop would move To E and catch the robber. So the cop would alternate between A and C waiting for the robber to go to a point adjacent to only E or only D and this would generate and infinite series that sums to less than 3.

Now if the cop follows the normal strategy he will move to B then the robber will move to a vertex adjacent only to D and then the cop will move to C the robber will have to move to d and then the cop will move to C catching the robber in three moves. So here the winning strategy is different from the optimal winning strategy is different from the optimal random strategy.

Answer by Kristal Cantwell for Generalizations of the Birkhoff-von Neumann Theorem

2009-11-26T19:51:42Z

I have found a paper on a generalization of the Birkhoff-von Neumann theorem here:

http://cowles.econ.yale.edu/conferences/2009/sum-09/theory/che.pdf

The authors are Eric Budish, Yeon-Koo Che, Fuhito Kojima, and Paul Milgram.

Here is the Abstract:

The Birkhoff-von Neumann Theorem shows that any bistochastic matrix can be written as a convex combination of permutation matrices. In particular, in a setting where n objects must be assigned to n agents, one object per agent, any random assignment matrix can be resolved into a deterministic assignment in accordance with the specified probability matrix. We generalize the theorem to accommodate a complex set of constraints encountered in many real-life market design problems. Specifically, the theorem can be extended to any environment in which the set of constraints can be partitioned into two hierarchies. Further, we show that this bihierarchy structure constitutes a maximal domain for the theorem, and we provide a constructive algorithm for implementing a random assignment under bihierarchical constraints. We provide several applications, including (i) single-unit random assignment, such as school choice; (ii) multi-unit random assignment, such as course allocation and fair division; and (iii) two- sided matching problems, such as the scheduling of inter-league sports matchups. The same method also finds applications beyond economics, generalizing previous results on the minimize makespan problem in the computer science literature

I have also found a master's thesis that involved generalization from a matrix to a hyper matrix, a matrix in higher dimensions. So one example would be a cubic array of numbers instead of a square. He proves a generalization to the three dimensional matrices which are called blocks. There are some open questions there as well. I found it interesting as I have wondered about extending the two dimensions of matrices to three coordinates and seeing what happened. It is available here:

https://ritdml.rit.edu/dspace/bitstream/1850/5967/1/NReffThesis05-18-2007.pdf

Answer by Kristal Cantwell for Reasons for the importance of planarity and colorability?

2011-01-21T18:05:33Z

There is the circle packing theorem, every connected simple planar graph is isomorphic to a circle packing.

Answer by Kristal Cantwell for Probability that a number and its digit reversal are relatively prime

2010-12-28T20:01:33Z

I think there are bases $b$ where the probability becomes arbitrarily low. Let $b$ be the product of the first $n$ primes plus one then the difference of $b$ and its palindrome will be divisible by $b-1$ and for them to be relatively prime neither can be divided by the first $n$ primes. So if $n$ is chosen large enough the probability can be made arbitrarily low.

Answer by Kristal Cantwell for Most intricate and most beautiful structures in mathematics

2010-12-12T19:04:42Z

The monster vertex algebra.

It is (to date) the central object in monstrous moonshine, since its character is the $q$-expansion of the modular $J$-function, its automorphism group is the monster simple group, and the graded trace of any element of the monster is the $q$-expansion of a genus zero modular function. The construction of this structure (by Frenkel, Lepowsky, and Meurman) involves ascending a hierarchy of objects that are by themselves quite intricate and beautiful.

One begins with the extended binary Golay code of length 24. Up to symmetries, it is the unique copy of $\mathbb{F}_2^{12}$ in $\mathbb{F}_2^{24}$ , for which any five basis vectors are contained in a unique codeword (i.e., it forms a Steiner $(5,8,24)$ system). The codewords are separated by Hamming distance at least 8, so even if 3 bits in a code word are changed the error can be corrected. The automorphism group of the Golay code is the sporadic simple group $M_{24}$ of order 244823040.
Using the Golay code to produce coordinates of generators, one constructs the Leech lattice $\Lambda$, which is a rather densely packed copy of $\mathbb{Z}^{24}$ in $\mathbb{R}^{24}$. One can also make the Leech lattice as a subquotient of the even unimodular lattice $I\!I_{25,1}$ , which has its own exceptional properties. Peter Shor mentioned the Leech lattice in another answer, so I'll just note that its automorphism group is a double cover of Conway's sporadic simple group $Co_1$, which has order 4157776806543360000.
For any positive definite even lattice $L$, there is a canonical construction of a vertex operator algebra graded by that lattice, called the lattice vertex algebra $V_L$. I think physicists say that it is the algebra of chiral symmetries of a conformal field theory describing a bosonic string propagating in the torus $L \otimes \mathbb{R}/L$ (but I may have mixed up the words). It has an action of the holomorph of the algebraic torus $L \otimes \mathbb{C}^\times$.
The "-1" automorphism of the Leech lattice induces an automorphism $\theta$ of $V_\Lambda$, and there is a unique irreducible $\theta$-twisted module $V_\Lambda(\theta)$ that inherits an action of the centralizer $2^{1+24}.Co_1$ of $\theta$ in the automorphism group of $V_\Lambda$. The monster vertex algebra is formed by taking the direct sum of fixed points: $V^\natural = (V_\Lambda)^\theta \oplus (V_\Lambda(\theta))^\theta$.

Apparently, the hard part was proving that the monster acts on $V^\natural$ by automorphisms.

There are some additional conjectural reasons for considering it beautiful:

In the same paper where it was constructed, it was conjectured to be the unique vertex operator algebra with central charge 24, character equal to the modular $J$ function, and representation category equivalent to $Vect$. (Naturally, this does not account for higher structure like twisted modules.)
Witten suggested that it is dual to pure 3-dimensional quantum gravity with minimal cosmological constant by AdS/CFT correspondence.

Answer by Kristal Cantwell for William Rowan Hamilton and Algebra as Time

2010-12-06T16:44:58Z

I believe that he may have been influenced by Kant. According to a quote from this website: "Kant maintains that geometry discovers the universal laws of space, and algebra discovers the universal laws of time. Space and time are "pure intuitions" by which perception can take place, so they are a priori and universal."

Answer by Kristal Cantwell for What is the current status of Agrawal's conjecture?

2010-11-21T19:59:46Z

I found a paper here: http://eprint.iacr.org/2009/008.pdf which generalizes a result from Lenstra's and Pomerance's paper.

The paper is "A note on Agrawal conjecture" by Roman Popovych.

Here is the abstract:

We prove that Lenstra proposition suggesting existence of many counterexamples to Agrawal conjecture is true in a more general case. At the same time we obtain a strictly ascending chain of subgroups of the group (Zp[X]/(Cr(X)))* and state the modified conjecture that the set {X-1, X+2} generate big enough subgroup of this group.

Here is the url for a paper from a student scientific conference containing some numerical results:

http://www.fmph.uniba.sk/fileadmin/user_upload/editors/studium/svk/2009/INF/vana.pdf.

Chromatic number of graphs of tangent closed balls

2009-12-08T19:11:40Z

The Koebe–Andreev–Thurston theorem gives a characterization of planar graphs in terms of disjoint circles being tangent. For every planar graph $G$ there is a disk packing whose graph is $G$. What happens when disks are replaced by closed balls? By closed balls of higher dimension? I have already asked one question about this here:

http://mathoverflow.net/questions/8031/graphs-of-tangent-spheres

The question I want to ask here is what is known about the chromatic numbers of these graphs? I have updated the numbers and changed the arguments in the following based on some of the answers.

Assume the chromatic number is 14 or more and we have the smallest such graph that is colorable with 14 or more colors. Take one of the smallest closed balls then since the kissing number for three dimensions is 12 there are at most 12 closed balls tangent to this closed ball. Remove this closed ball then the remaining graph can be colored in 13 or less colors. Color it with 13 colors. Then add the closed ball back in since it is tangent to only 12 closed balls it can be given one of the thirteen colors so we have the entire graph can be colored with thirteen colors which gives a contradiction so the chromatic number must be 13 or less. We have an lower bound of 6 from a spindle constructed according to David Eppstein's answer. Can we improve on the 6 to 13 range?

We have the lower bound is a quadratic function and we have an upper bound that is exponential. Which of these two is right?

Is there a case where closed balls of different sizes raise the chromatic number from closed balls the same size?

Finally based on the existing chromatic numbers I am wondering if it is possible to answer this question. Is there a dimension where the chromatic number of the unit distance graph is different from the chromatic number of the graphs in that dimension of tangent closed balls. The unit distance graph is the set of all points in the $n$-dimensional space with two points connected if their distance is one. For dimension two the chromatic number is known to be in the range from 4 to 7. For dimension three the range is 6 to 15. For the graphs of tangent disks we have a chromatic number of 4 and for closed balls a range from 6 to 13. So the possibility that the chromatic numbers of the two types of graphs are the same has not yet been eliminated. So the specific question is what is known and what can be proved about the chromatic number of the graphs of tangent closed balls?

Answer by Kristal Cantwell for Fundamental Examples

2009-11-22T21:36:49Z

The Navier-Stokes equations.

Answered by Kristal Cantwell

Answer by Kristal Cantwell for How many orders of infinity are there?

2010-06-26T18:00:51Z

I think that there a complete set of growth functions of intermediate cardinality. This is based a earlier discussion on a related subject here. In particular Joel David Hamkins answer seems to answer the question in the affirmative.

Answer by Kristal Cantwell for When Have Numerology and Computational Experimentation Been Successful?

2010-06-18T22:44:54Z

There is this thread about experimental mathematics.

Answer by Kristal Cantwell for Mathematics and autodidactism

2010-06-17T16:53:32Z

I don't agree that it is always easier to learn from people than from books. I taught myself a lot of mathematics without instruction before college. I think in some cases interacting with people is useful because it is easier for them to see mistakes you are making. But that could be done via the internet. I think for some people social interaction might be needed to learn. The ideal learning situation might vary from person to person.

A Counterexample to the HIrsch Conjecture

2010-05-28T22:07:30Z

Recently Francisco Santos has announced that he has a counterexample to the Hirsch conjecture. The last I heard it was circulating among several people and there would be a public version of it available soon. I am curious how close it is to release. Also has there been any progress in the attempt to find the vertices of the counterexample. The last I heard to find the vertices a series of steps had to be done and each step increased the complexity of the problem by a geometric factor making it difficult to complete the computation.

Answer by Kristal Cantwell for A Counterexample to the HIrsch Conjecture

2010-06-15T18:02:15Z

The public version is now out. It is available here

Comment by Kristal Cantwell

2012-10-27T07:35:34Z

arxiv.org/pdf/1209.4986.pdf

Comment by Kristal Cantwell

2012-09-26T17:37:21Z

Could you look at areas? If when half circles meet nicely the area in common was always greater than a quarter of the area of a full circle than any three half circles that meet nicely in pairs would have a point in common.

Comment by Kristal Cantwell

2012-04-10T17:23:30Z

The McKay paper is available [here][1] [1]: cs.anu.edu.au/~bdm/papers/rt.pdf

Comment by Kristal Cantwell

2011-07-13T17:16:55Z

Yes there was a mistake there, I think I corrected it.

Comment by Kristal Cantwell

2011-03-28T03:55:32Z

I have fixed the link.

Comment by Kristal Cantwell

2011-02-11T19:55:16Z

I have replaced the URL and added the names of the authors.

Comment by Kristal Cantwell

2010-12-06T20:26:56Z

Quaternions can be used to describe rotations in three dimensional space in which the real term is the cosine of half the angle and the other three coordinates represent the axis of rotation so that could be motivation for thinking of three of the coordinates spatial and the fourth as non-spatial.

Comment by Kristal Cantwell

2010-12-05T17:53:41Z

There is a stack exchanged devoted to TeX at tex.stackexchange.com maybe you could try this question there.

Comment by Kristal Cantwell

2010-11-14T17:30:36Z

Rubber bands, convex embeddings and graph connectivity by N. Linial, L. Lovász and A. Wigderson may be the Lovasz paper mentioned above.

Comment by Kristal Cantwell

2010-07-15T07:55:22Z

Doesn't the point (3,0,5,5,5,5) which is blue in this coloring have red neighbors (3,1,5,5,5,5), (3,-1,5,5,5,5,), (2,0,5,5,5.5) and (4,0,5,5,5,5) thus making the sensitivity at least 4?

Comment by Kristal Cantwell

2010-06-23T04:48:35Z

There is still a problem with the averaging argument. The proof as it is doesn't work.

Comment by Kristal Cantwell

2010-06-23T04:06:51Z

I have corrected the sentence to read "Because of this the average half circle must contain parts of at least $n$ or less arcs." I am counting arcs that are in ore are partly in.

Comment by Kristal Cantwell

2010-06-22T16:56:47Z

I rewrote the averaging argument.

Comment by Kristal Cantwell

2010-06-22T06:08:19Z

That should have been $n$ arcs, I have corrected it.

Comment by Kristal Cantwell

2010-06-13T06:20:55Z

They are closer to each other than they are to the nearest integer at least by a factor of 10 perhaps more I can't really tell since the expansion stops before they differ.